When an object is placed 40mathrm~cm$40\mathrm{~cm}$ away from a spherical mirror an image of magnification frac12$\frac{1}{2}$ is produced. To obtain an image with magnification of frac13$\frac{1}{3}$, the object is to be moved:

### Related Formula The magnification (m$m$) of a spherical mirror is given by: m = fracff - u$$m = \frac{f}{f - u}$$ where, f$f$ = focal length (R/2$R/2$) u$u$ = object distance ### Core Logic Given radius of curvature, R = 30 mathrm~cm$R = 30 \mathrm{~cm}$. For a convex mirror: f = +fracR2 = +15 mathrm~cm$$f = +\frac{R}{2} = +15 \mathrm{~cm}$$ Since a convex mirror always forms a virtual, erect, and diminished image of a real object, the magnification m$m$ must be positive: m = +frac12$$m = +\frac{1}{2}$$

### Step 1: Solve for Object Distance Using the magnification formula: +frac12 = frac1515 - u$$+\frac{1}{2} = \frac{15}{15 - u}$$ 15 - u = 30 implies u = -15 mathrm~cm$$15 - u = 30 \implies u = -15 \mathrm{~cm}$$ Thus, the object distance is -15 mathrm~cm$-15 \mathrm{~cm}$. ### Pattern Recognition Remember: Convex mirrors produce *only* virtual images for real objects, which means m$m$ is always positive and less than 1$1$. If the question mentioned a concave mirror with a diminished image of half size, the image would be real, and m$m$ would be negative. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments

### Related Formula From the Lens Maker's Formula: frac1f = left( fracmu_textlensmu_textmedium - 1 right) left( frac1R_1 - frac1R_2 right)$$\frac{1}{f} = \left( \frac{\mu_{\text{lens}}}{\mu_{\text{medium}}} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$$ Taking the ratio of focal length in liquid medium (f_m$f_m$) to focal length in air (f_a$f_a$): fracf_mf_a = frac(mu_1 - 1) mu_mmu_1 - mu_m$$\frac{f_m}{f_a} = \frac{(\mu_1 - 1) \mu_m}{\mu_1 - \mu_m}$$ where, mu_1$\mu_1$ = refractive index of the lens material = 1.5 mu_m$\mu_m$ = refractive index of the liquid medium = 1.6 f_a$f_a$ = focal length in air = 20 mathrm~cm$20 \mathrm{~cm}$ ### Core Logic Substitute the parameters into the relative ratio template equation: fracf_m20 = frac(1.5 - 1) times 1.61.5 - 1.6$$\frac{f_m}{20} = \frac{(1.5 - 1) \times 1.6}{1.5 - 1.6}$$ ### Step 1: Simplify and Compute $fracf_m20 = frac0.5 times 1.6-0.1$$$\frac{f_m}{20} = \frac{0.5 \times 1.6}{-0.1}$$ fracf_m20 = frac0.8-0.1 = -8$$\frac{f_m}{20} = \frac{0.8}{-0.1} = -8$$ f_m = -8 times 20 = -160 mathrm~cm$$f_m = -8 \times 20 = -160 \mathrm{~cm}$$ Therefore, the focal length in the liquid is -160 mathrm~cm$-160 \mathrm{~cm}$. ### Pattern Recognition Notice that since the surrounding liquid medium has a higher refractive index than the lens material itself (mu_m gt mu_1$\mu_m \gt \mu_1$), the sign of the focal length flips from positive to negative. The convex lens behaves as a diverging lens inside this specific liquid. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments

### Related Formula textNewton's Lens Formula: x_1 x_2 = f^2$$\text{Newton's Lens Formula: } x_1 x_2 = f^2$$ where x_1$x_1$ and x_2$x_2$ are object and image distances from the focal points. ### Core Logic Since distances x$x$ and y$y$ are measured from the focal point (not the optical center), we use Newton's formula: x y = f^2$x y = f^2$. From the graph, a prominent point on the curve is (20, 20)$(20, 20)$. ### Step 1: Calculate Focal Length 20 times 20 = f^2$$20 \times 20 = f^2$$ f^2 = 400 implies f = 20 mathrm~cm$$f^2 = 400 \implies f = 20 \mathrm{~cm}$$ Alternatively using standard lens formula: Object distance from optical center u = -(f + x)$u = -(f + x)$ Image distance v = +(f + y)$v = +(f + y)$ frac1v - frac1u = frac1f$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$ frac1f+y - frac1-(f+x) = frac1f$$\frac{1}{f+y} - \frac{1}{-(f+x)} = \frac{1}{f}$$ If x = y = 20 mathrm~cm$x = y = 20 \mathrm{~cm}$: frac1f+20 + frac1f+20 = frac1f$$\frac{1}{f+20} + \frac{1}{f+20} = \frac{1}{f}$$ frac2f+20 = frac1f$$\frac{2}{f+20} = \frac{1}{f}$$ 2f = f + 20 implies f = 20 mathrm~cm$$2f = f + 20 \implies f = 20 \mathrm{~cm}$$ ### Pattern Recognition Whenever distances are specified relative to the focal point, Newton's formula (xy = f^2$xy = f^2$) instantly solves the problem. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments

### Related Formula m = fracvu$$m = \frac{v}{u}$$ frac1f = frac1v - frac1u$$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$$ ### Core Logic For a real image produced by a convex lens, the magnification m$m$ is negative. The distance between the object and the real image is the absolute sum of their distances from the lens: |v| + |u| = v - u = 45 mathrm~cm$|v| + |u| = v - u = 45 \mathrm{~cm}$ (since u$u$ is negative and v$v$ is positive). ### Step 1: Set Up Magnification and Distance fracvu = -2$$\frac{v}{u} = -2$$ v = -2u quad dots (i)$$v = -2u \quad \dots (i)$$ Distance between object and image: v - u = 45 quad dots (ii)$$v - u = 45 \quad \dots (ii)$$ ### Step 2: Solve for Object and Image Distances Substitute (i) into (ii): (-2u) - u = 45$(-2u) - u = 45$ -3u = 45 Rightarrow u = -15 mathrm~cm$$-3u = 45 \Rightarrow u = -15 \mathrm{~cm}$$ Then, v = -2(-15) = +30 mathrm~cm$v = -2(-15) = +30 \mathrm{~cm}$. ### Step 3: Calculate Focal Length frac1f = frac1v - frac1u$$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$$ frac1f = frac130 - frac1-15$$\frac{1}{f} = \frac{1}{30} - \frac{1}{-15}$$ frac1f = frac130 + frac230 = frac330 = frac110$$\frac{1}{f} = \frac{1}{30} + \frac{2}{30} = \frac{3}{30} = \frac{1}{10}$$ f = +10 mathrm~cm$$f = +10 \mathrm{~cm}$$ ### Pattern Recognition For a real image, total object-to-image distance D = v + |u|$D = v + |u|$. Applying sign convention naturally resolves this to D = v - u$D = v - u$. Two conditions (magnification + total distance) reliably solve for both u, v$u, v$ before hitting the lens equation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments

### Related Formula fracmu_2v - fracmu_1u = fracmu_2 - mu_1R$$\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$$ ### Core Logic For a convex refracting surface, the radius of curvature R$R$ is positive if the center of curvature lies in the denser medium. mu_1 = 1 text (air) $ \mu_1 = 1 \text{ (air)} $ mu_2 = 1.5 text (glass/medium) $ \mu_2 = 1.5 \text{ (glass/medium)} $ u = -100 text cm $ u = -100 \text{ cm} $ R = +20 text cm $ R = +20 \text{ cm} $

### Step 1: Calculate Image Position frac1.5v - frac1-100 = frac1.5 - 120$$\frac{1.5}{v} - \frac{1}{-100} = \frac{1.5 - 1}{20}$$ frac1.5v + frac1100 = frac0.520$$\frac{1.5}{v} + \frac{1}{100} = \frac{0.5}{20}$$ frac1.5v = frac140 - frac1100$$\frac{1.5}{v} = \frac{1}{40} - \frac{1}{100}$$ frac1.5v = frac5 - 2200 = frac3200$$\frac{1.5}{v} = \frac{5 - 2}{200} = \frac{3}{200}$$ v = frac1.5 times 2003 = 100 text cm$$v = \frac{1.5 \times 200}{3} = 100 \text{ cm}$$ ### Step 2: Distance from Object The image is formed at v = +100 text cm$v = +100 \text{ cm}$ from the pole, which is on the other side of the surface. The object is at |u| = 100 text cm$|u| = 100 \text{ cm}$ from the pole. Distance between object and image = |u| + v = 100 + 100 = 200 text cm$= |u| + v = 100 + 100 = 200 \text{ cm}$. ### Pattern Recognition Always read the final line of optics questions carefully. The question asks for distance "from the object" not "from the surface/pole". This is a classic trap where students answer 100 instead of 200. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments